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(A) To find the area $A$,we first determine the intersection points of the curves $y = x^2+2$ and $y = 2x+2$.Setting $x^2+2 = 2x+2$,we get $x^2 - 2x =

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$164$

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(A) To find the area $A$,we first determine the intersection points of the curves $y = x^2+2$ and $y = 2x+2$.Setting $x^2+2 = 2x+2$,we get $x^2 - 2x = 0$,which implies $x(x-2) = 0$. Thus,the curves intersect at $x=0$ and $x=2$.For $0 \leq x \leq 2$,we have $x^2+2 \leq 2x+2$,so $\min \{x^2+2, 2x+2\} = x^2+2$.For $2 \leq x \leq 3$,we have $2x+2 \leq x^2+2$,so $\min \{x^2+2, 2x+2\} = 2x+2$.The area $A$ is given by:$A = \int_0^2 (x^2+2) dx + \int_2^3 (2x+2) dx$Calculating the integrals:$\int_0^2 (x^2+2) dx = [\frac{x^3}{3} + 2x]_0^2 = (\frac{8}{3} + 4) - 0 = \frac{20}{3}$$\int_2^3 (2x+2) dx = [x^2 + 2x]_2^3 = (9+6) - (4+4) = 15 - 8 = 7$Thus,$A = \frac{20}{3} + 7 = \frac{20+21}{3} = \frac{41}{3}$.Finally,$12A = 12 	imes \frac{41}{3} = 4 	imes 41 = 164$.

Let the area of the region $\{(x, y): 0 \leq x \leq 3, 0 \leq y \leq \min \{x^2+2, 2x+2\}\}$ be $A$. Then $12A$ is equal to

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